Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $k = \dfrac{-4r - 28}{-10r - 30} \div \dfrac{r^2 + 8r + 7}{r + 3} $
Dividing by an expression is the same as multiplying by its inverse. $k = \dfrac{-4r - 28}{-10r - 30} \times \dfrac{r + 3}{r^2 + 8r + 7} $ First factor the quadratic. $k = \dfrac{-4r - 28}{-10r - 30} \times \dfrac{r + 3}{(r + 7)(r + 1)} $ Then factor out any other terms. $k = \dfrac{-4(r + 7)}{-10(r + 3)} \times \dfrac{r + 3}{(r + 7)(r + 1)} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac{ -4(r + 7) \times (r + 3) } { -10(r + 3) \times (r + 7)(r + 1) } $ $k = \dfrac{ -4(r + 7)(r + 3)}{ -10(r + 3)(r + 7)(r + 1)} $ Notice that $(r + 3)$ and $(r + 7)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac{ -4\cancel{(r + 7)}(r + 3)}{ -10(r + 3)\cancel{(r + 7)}(r + 1)} $ We are dividing by $r + 7$ , so $r + 7 \neq 0$ Therefore, $r \neq -7$ $k = \dfrac{ -4\cancel{(r + 7)}\cancel{(r + 3)}}{ -10\cancel{(r + 3)}\cancel{(r + 7)}(r + 1)} $ We are dividing by $r + 3$ , so $r + 3 \neq 0$ Therefore, $r \neq -3$ $k = \dfrac{-4}{-10(r + 1)} $ $k = \dfrac{2}{5(r + 1)} ; \space r \neq -7 ; \space r \neq -3 $